Power banks in parallel to increase A / mAh

[ericwilk]

Hi there,

I am trying to power a device that will draw ~1-1.5A (2 peak at most I'd think) that I want to be portable. It needs to fit in a chassis and I found some cheap tiny power banks but they are rated at 1A and last 2200 mAh. I was wondering if like batteries I could hook them in parallel to get more out of them but retain the 5V.

To protect the units and help evenly spread the current (at least I think this might work) I was thinking of adding a few diodes and a cap where they join, something like this:

Is this feasible?

Also, I have 5V voltage regulators on order but idk if that's totally necessary.

Push comes to shove, I just went at my laser printer with a power drill when it pissed me off and there's a board with damn near everything. It also said "don't open -- laser -- may cause blindness" so I thought "COOL" and I have a frickin' laser beam now. An IR one I can't use for anything (well, unless I was into B&E and wanted to fool motion sensors but that's not my thing.)

Pardon my ignorance, but I haven't dealt with this stuff for a while. Did a search but couldn't find much relevant.




TIA,

~ Eric

 

[Arouse1973]

What type of battery is it you are trying to join together? If they are primary cells like lithium Thionyl chloride then yes you will need diode if connecting more than two cells in parallel. The diodes need to have a very low reverse leakage current which conforms to UL regulations. If they are rechargeable lithium ion packs already qualified then you don't need the diodes but they must be matched packs and be of the same charge level when you connect them preferably 20% charged. If you are going to charge them together then I recommend using a charger of 1C max if the batteries allow 1C charge rate but don't go any higher.
Adam

 

[ericwilk]

Specifically, they are these: http://www.radioshack.com/radioshack-2200mah-portable-power-bank-white/2302016.html#.VM0X84Xk5hE

It didn't say on them but according to the user's guide they are Li-ion.

Thanks a bunch!

 

[BobK]

That unit will consist of a LiIon battery (3.7V) and a boost converter to up it to 5V.

Paralleling them would not work well. If you want more power, buy a larger capacity power pack.

Bob

 

[ericwilk]

Darn. I have larger power packs here. The problem is getting them to fit inside the chassis.

Maybe I can just use rechargeable batteries and downconvert >.>

 

[Arouse1973]

You could maybe try putting them in series and use a buck. But you will only have the capacity of one cell. Although initially you would draw less average current from the battery. But as the voltage dropped this would increase and not give you much of a benefit.
Adam

 

[BobK]

Not true. A perfectly efficient buck converter would give you the sum of the capacities if you placed the batteries in series. A realistic one would give you 80% of the sum of the capacities.


Bob

 

[ericwilk]

I'm looking for current, not potential.

I still want 5V, just more mAh / peak amperage.

 

[Arouse1973]

Not true. A perfectly efficient buck converter would give you the sum of the capacities if you placed the batteries in series. A realistic one would give you 80% of the sum of the capacities.


Bob

I am doing a test at the moment with two series 2250mA batteries and a buck drawing 25mA. So your saying I should get over 4Ah of capacity then? I let you know how I got on.
Adam

 

[ericwilk]

I am doing a test at the moment with two series 2250mA batteries and a buck drawing 25mA. So your saying I should get over 4Ah of capacity then?

No, I was thinking in series it would ~ double the voltage.

Correct me if I'm wrong, here...

 

[Gryd3]

No, I was thinking in series it would ~ double the voltage.

Correct me if I'm wrong, here...

Series would indeed increase the voltage, but by using a 'buck' type converter, the voltage would be brought back down to a regulated 5V. This 'Buck' converter works similar to a transformer in that the power going out will be roughly the same power going in.
So putting 10V 1A in, would allow 5V and ~1.6A out (depending on the efficiency of the buck converter).
The tricky part with the power banks you are wanting to use is that the output is from a 'boost' switch-mode power supply. Which means that the circuitry actually pulses the output and sends it though some additional components to boost the voltage up to 5V. Connecting these devices in parallel can cause some... odd results, and is not recommended.

Ideally, you best solution would be to visit eBay or another source and find a 5V portable source that will provide the output current you require.
It's that, or buy your own components and make your own. (Battery, Charge controller, Boost (or Buck) converter, old USB extension cable or USB hub to salvage a USB end off of)

 

[ericwilk]

Hmm... Interesting. Thanks!

I'll have to look into that.

I can always hook it up to a meter and monitor it, or even simulate ~1.5A using resistance or whatever and use a potentiometer to see what happens under a curve of steady current and spikes, etc.

I have a way to charge it via USB but I want it to be portable and self-contained.

Regular rechargeable batteries might work, though along with a regulator (might not even be needed; aren't rechargeable batteries something like 1.2 or 1.3V and not 1.5? I have like 10 5V regulators coming in the mail, though.)

Thanks again for everything! Just dealing with some sensitive equipment is all so I thought I'd ask people who know more than me about this stuff.

 

[BobK]

I am doing a test at the moment with two series 2250mA batteries and a buck drawing 25mA. So your saying I should get over 4Ah of capacity then? I let you know how I got on.
Adam

If your two batteries are 5V 1000mAH, giving you 10V out at 2000mAH, and your buck converter goes back down to 5V, at 100% efficiency, it would be the equivalent of 2000mAH. Why? Because the buck converter from 10V to 5V have a 50% duty cycle drawing only the current needed by the load. On the other half the current is provided by the inductor. As Grd3 noted, It will really be more like 1.6 to 1.8 times the capacity.

Another way to look at it: Say the load draws 1A. If you draw 1A from the 2 series batteries at 10V that is twice the power needed by the device using 5V at 1A. The rest of that power is stored in the inductor, and released while the batteries sit there and rest half the time.

Edit: That is why buck converters are preferred to linear regulators.

Bob

 

[ericwilk]

Thanks! Just never used one so I'll need to do a bit of reading.

Unfortunately that won't be until I complete something in the professional world...

I'm a software developer and looking at a place that wants to hire me as a senior developer (big company too, and very successful. also nearby) but I have a project I need to complete before the in-person interview (that would be the 3rd round).

Soooooo yeah, I want to get that done by Monday. There's another company I like, a little out of the way but getting a car isn't out of the question since I don't live right in the city, anyway (I did for 6 years and living and working in SF got to be a bit much. I still love the place, though.) The bigger company would likely pay me a lot more, though >.>

Decisions, decisions...

 

[Arouse1973]

If your two batteries are 5V 1000mAH, giving you 10V out at 2000mAH, and your buck converter goes back down to 5V, at 100% efficiency, it would be the equivalent of 2000mAH. Why? Because the buck converter from 10V to 5V have a 50% duty cycle drawing only the current needed by the load. On the other half the current is provided by the inductor. As Grd3 noted, It will really be more like 1.6 to 1.8 times the capacity.

Another way to look at it: Say the load draws 1A. If you draw 1A from the 2 series batteries at 10V that is twice the power needed by the device using 5V at 1A. The rest of that power is stored in the inductor, and released while the batteries sit there and rest half the time.

Edit: That is why buck converters are preferred to linear regulators.

Bob

Yes I agree with this as far as if the voltage in is twice the voltage out and stays there. But when the voltage falls with a battery the gain in relative capacity is reduced. Here is my simulation of what I am doing at the moment. You can see the average current over the discharge cycle is 80% of the load. 3.6V out 10mA load




And this is for 5V out 10mA load.


Thanks
Adam

 

[ericwilk]

It might not be THAT much of a big deal if it's like... a half hour less time, TBH

I'm going with your suggestion, though. I'm just in the middle of some work stuff.

I have a few inductors I got off circuit boards and from cables but I need to do the math to figure out what is good.

Or, it seems from reading (more like skimming) a bit, it just has to be able to hold enough in the field. These look like GOOD inductors, so that may just work. Or wrap a wire around a magnetic loop.

Also, I realized the crappy transistors I bought have a rating of 600 mA?

So, I need to go to Radio Shack to get better ones, or at least a de-soldering iron.

I have a reflow station but I keep burning myself when the board gets hot and I hit the wrong spot (and after ripping off 10-20 components it does; probably not good for the components, either). The solder-sucker I got never worked really well. I like the all-in-one kind with the bulbs to get bottom-soldered stuff.

 

[Arouse1973]

Thanks! Just never used one so I'll need to do a bit of reading.

Unfortunately that won't be until I complete something in the professional world...

I'm a software developer and looking at a place that wants to hire me as a senior developer (big company too, and very successful. also nearby) but I have a project I need to complete before the in-person interview (that would be the 3rd round).

Soooooo yeah, I want to get that done by Monday. There's another company I like, a little out of the way but getting a car isn't out of the question since I don't live right in the city, anyway (I did for 6 years and living and working in SF got to be a bit much. I still love the place, though.) The bigger company would likely pay me a lot more, though >.>

Decisions, decisions...

Good Luck